Probability of drawing a full house

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Discussion Overview

The discussion revolves around calculating the probability of drawing a full house (three of a kind and two of a kind) from a standard deck of 52 cards without replacement. Participants explore the combinatorial methods involved in determining this probability, including the selection of card types and the application of combinatorial coefficients.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant seeks guidance on how to approach the probability calculation for drawing a full house, indicating uncertainty about the method.
  • Another participant suggests a counting method involving selecting a card type for the three of a kind and then for the two of a kind, emphasizing the multiplication of choices.
  • A participant expresses understanding of the selection process but struggles with the specifics of calculating the number of ways to pick the required cards.
  • Clarifications are provided regarding the combinatorial coefficients for selecting cards, specifically C(4,3) for three cards and C(4,2) for two cards.
  • There is a discussion about the total number of possible hands being represented by C(52,5), which is the total combinations of drawing five cards from the deck.
  • Participants discuss the reasoning behind multiplying the number of ways to choose card types and the number of ways to choose specific cards from those types.
  • A participant raises a related question about calculating the probability of guessing answers correctly on a multiple-choice test, comparing two different probability approaches.
  • Clarification is given regarding the difference between calculating the probability of guessing the first three questions correctly versus guessing exactly three correctly.

Areas of Agreement / Disagreement

Participants generally agree on the methods for calculating the probability of drawing a full house, but there is some uncertainty and confusion regarding the application of combinatorial principles and the interpretation of probability in different contexts.

Contextual Notes

Participants reference combinatorial coefficients and the independence of events in probability, but there are unresolved questions about the application of these concepts in specific scenarios.

Who May Find This Useful

This discussion may be useful for students studying probability and combinatorics, particularly in the context of card games and related mathematical problems.

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Homework Statement


Whats the probability of drawing a full house(three of a kind and two of a kind) from a deck of 52 cards without replacement.


Homework Equations





The Attempt at a Solution


I know that the first two cards are fine but next one should either be of the first one drawn or the second one drawn. Then the last two should be of the other kind that wasnt matched previously but I don't know how to look at that in terms of probability. How should I go about doing such a question. Got this in my Mech Eng Lab I class.
 
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There's an easier way to count them. Pick a card type for the three of a kind. There are 13 ways to do that. Now how many ways to pick three cards for each card type? There are 12 choices for the second card type. Now how many ways to pick two cards from each card type? Multiply those all together and you have the total number of possible full house hands. Now divide by the total number of possible hands.
 
I understand the 13 and 12 part as there are 13 sets in the 52 card pack and if you pick one kind, you'll clearly have 12 sets that your are able to pick from the second time around. But I am still at a lost. I really don't know how many ways to pick three cards for each card type as well as the latter. *sighs*
 
Ok. How many ways to pick 3 is easy. Of the four cards you can leave anyone out. There are four ways to do that. This is often written C(4,3)=4. Look up 'combinatorial coefficient'. Similarly the number of ways to choose 2 from 4 is C(4,2)=6. C(n,k) is n!/(k!*(n-k)!). Does that seem familiar? You'll need the general formula to figure out how many ways to choose 5 cards from 52. It's C(52,5). Can you work it out?
 
yeah all those formulas have been used by me at least once between college and primary school(junior high i guess for you guys). So now i understand how many ways to pick 3 cards and as a result how to do two as well. Now the result of say 4nCr3. What do i do with it. You said multiply it by 4nCr2. Why was that done? Then you said put it over the total number of possible hands. What would that be? the total number of drawing five cards or something...? Forgive the ignorance please.
 
13*C(4,3)*12*C(4,2)/C(52,5) is the answer. Can you tell me why that works in words? I'll start you off. C(52,5) is the total number of possible hands (52 cards, choose 5).
 
Oh yes. High school probability is coming back to me. I knew that last part. From you told me earlier that total it should be divided by the total possible outcomes, I figured it would be 52nCr5 as and that will give you the total combination of five cards in general, whether it be full house or not. As for the mulitiplying them, I am not sure why you did that. I remember that we had used this rule that AND was mulitiply and OR was addition. The former applies here cause they're independent if you adjust the cards remaning accordingly. I just need to iron out why it was required to multiply the 13 and 12 in. What does that rule state?
 
First I pick 1 type out of 13 AND then I pick one type out of the remaining 12. I have to do both. The same for the number of cards.
 
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No no I know why you multiplied each other because they're independent events. But I got it nevertheless after reflecting on the matter. You multiplied the first one by 13 because after finding the ways of drawing three cards from four cards(a set) you needed to find how many would come from 13 sets which would require the multiplication and so and so forth for the other.

That though brings me to another question.

Say I am doing a multiple choice paper of 5 questions and I ask what's the probability of guessing the first 3 correctly. Since the probability of getting a question right is 1/5, would the probability of guessing three right by (1/5)^3 = 1/125? Or would i have to use the formula P(n) = (N!/((N-n)!N!))*(P^n)*(1-P)^(N-n)?
 
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  • #10
If the question asks for only the first three correctly (and doesn't care about the other two). Then the first approach is fine. If it asks for the probability of guessing EXACTLY three correctly (and doesn't care which ones) then the second approach is fine.
 
  • #11
Okay thanks man. You have been a big help.
 

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